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Answer: The [tex]K_{sp}[/tex] of [tex]AgCrO_{4}[/tex] is [tex]1.1 \times 10^{-12}[/tex].
Explanation:
Given: [tex][Ag^{+}] = 1.3 \times 10^{-4} M[/tex]
The reaction equation will be written as follows.
[tex]Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+} + CrO^{2-}_{4}[/tex]
This shows that the concentration of [tex]CrO^{2-}_{4}[/tex] is half the concentration of [tex]Ag^{+}[/tex] ion. So,
[tex][CrO^{2-}_{4}] = \frac{1.3 \times 10^{-4}}{2}\\= 0.65 \times 10^{-4} M[/tex]
The expression for [tex]K_{sp}[/tex] of this reaction is as follows.
[tex]K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}][/tex]
Substitute values into the above expression as follows.
[tex]K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}]\\= (1.3 \times 10^{-4})^{2} \times 0.65 \times 10^{-4}\\= 1.1 \times 10^{-12}[/tex]
Thus, we can conclude that the [tex]K_{sp}[/tex] of [tex]AgCrO_{4}[/tex] is [tex]1.1 \times 10^{-12}[/tex].
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